Whakaoti i te 5x^2+12x-44=

$$5 x ^ { 2 } + 12 x - 44 =$$

Tauwehe

$$\left(x-2\right)\left(5x+22\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 5x^{2}+ax+bx-44. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=12 ab=5\left(-44\right)=-220$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -220.

$$-1,220 -2,110 -4,55 -5,44 -10,22 -11,20$$

Tātaihia te tapeke mō ia takirua.

$$-1+220=219 -2+110=108 -4+55=51 -5+44=39 -10+22=12 -11+20=9$$

Ko te otinga te takirua ka hoatu i te tapeke 12.

$$a=-10 b=22$$

Tuhia anō te 5x^{2}+12x-44 hei \left(5x^{2}-10x\right)+\left(22x-44\right).

$$\left(5x^{2}-10x\right)+\left(22x-44\right)$$

Tauwehea te 5x i te tuatahi me te 22 i te rōpū tuarua.

$$5x\left(x-2\right)+22\left(x-2\right)$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(5x+22\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$5x^{2}+12x-44=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$x=\frac{-12±\sqrt{12^{2}-4\times 5\left(-44\right)}}{2\times 5}$$

Pūrua 12.

$$x=\frac{-12±\sqrt{144-4\times 5\left(-44\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-12±\sqrt{144-20\left(-44\right)}}{2\times 5}$$

Whakareatia -20 ki te -44.

$$x=\frac{-12±\sqrt{144+880}}{2\times 5}$$

Tāpiri 144 ki te 880.

$$x=\frac{-12±\sqrt{1024}}{2\times 5}$$

Tuhia te pūtakerua o te 1024.

$$x=\frac{-12±32}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{-12±32}{10}$$

Nā, me whakaoti te whārite x=\frac{-12±32}{10} ina he tāpiri te ±. Tāpiri -12 ki te 32.

$$x=\frac{20}{10}$$

Whakawehe 20 ki te 10.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{-12±32}{10} ina he tango te ±. Tango 32 mai i -12.

$$x=-\frac{44}{10}$$

Whakahekea te hautanga \frac{-44}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{22}{5}$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te 2 mō te x_{1} me te -\frac{22}{5} mō te x_{2}.

$$5x^{2}+12x-44=5\left(x-2\right)\left(x-\left(-\frac{22}{5}\right)\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$5x^{2}+12x-44=5\left(x-2\right)\left(x+\frac{22}{5}\right)$$

Tāpiri \frac{22}{5} ki te x mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$5x^{2}+12x-44=5\left(x-2\right)\times \frac{5x+22}{5}$$

Whakakorea atu te tauwehe pūnoa nui rawa 5 i roto i te 5 me te 5.

$$5x^{2}+12x-44=\left(x-2\right)\left(5x+22\right)$$

Aromātai

$$\left(x-2\right)\left(5x+22\right)$$